The h Bar

0celo7

15:00

Did I have to find $k$ or something??

Or $C$

ok

ACuriousMind

@Sean Then you could try looking which pole of the electromagnet the compass points at, and which pole of it the bar magnets are attracted to. That might be more reliable than hanging a bar magnet from your ceiling and hoping the earth's magnetic field is strong enough to align it

0celo7

$y'=-2k/x^3, k=-x^3y'/2,x^2y=-x^3y'/2$

o.o

what is going on

$y'=-2y/x$

$\tilde y'=x/2y$

WHAT THE HELL

ACuriousMind

Dude, stop posting the same equation over and over again :P

0celo7

because I've done the problem 3 ways and keep getting the same equation ?????

but it's wrong

Bastian Treichler

normally, is a BSc doing his MSc a graduate or an undergraduate? or does that depend on the country/university?

0celo7

15:04

@BastianTreichler What?

Bastian Treichler

@0celo7 if someone has received his BSc, and then enrolls to do his MSc, is he considered an undergraduate student or a graduate?

0celo7

maybe I have to solve for $C$

how do I do this

@BastianTreichler If he has an undergraduate degree and is getting a graduate degree, I think he's a high schooler /s

I'm not sure what you're asking

yuggib

@BastianTreichler For me, grad student means PhD student

i.e. after MSc in most countries

0celo7

@yuggib You're strange.

yuggib

but that's only my opinion

0celo7

15:06

@yuggib You can skip the MS in some disciplines.

ACuriousMind

@BastianTreichler Being a "graduate student" means you are a student which already has a (prerequisite) degree, i.e. MSc students are technically graduate students

@0celo7 No, you normally can't in most European systems.

yuggib

@0celo7 I know, but I don't consider those disciplines

0celo7

@yuggib What the fuck?

Engineering is not a discipline?

How is this site not hostile to engineers?

Slereah

Sorry @0celo7

Engineering is more of a hobby

Fancy legos

yuggib

@0celo7 never been a worthwhile one, no

:P

0celo7

15:09

I don't know if you're kidding.

I suspect you aren't.

yuggib

:-D

Bastian Treichler

so it seems the definition does depend on the region. @yuggib where are you from?

yuggib

engineering is more useful than psychology...

0celo7

Seriously, how aren't the slope lines of $x^4/8k+C$ orthogonal to those of $k/x^2$

ACuriousMind

@Slereah Is Fallout 4 easily minimized, or do you have a second device/screen? The other Fallout/TES games never liked me minimizing them.

0celo7

15:10

^

yuggib

@BastianTreichler Italy, however I have been around in Europe

Slereah

Easily minimized

0celo7

@yuggib Yugoslavia?

Slereah

Although to be fair

I can always minimize a game

0celo7

@Slereah What hardware are you running?

Slereah

15:11

I have my ways

0celo7

GPU especially

yuggib

and if a book is for grad students, it is usually for PhD students/last year of MSc

0celo7

When I minimize Skyrim I freeze my computer.

Bastian Treichler

ok thanks for the answers @ all

Slereah

16 GB RAM, 4 GHz CPU

0celo7

15:11

GPU

Slereah

Looking for it

ACuriousMind

@all really should ping everyone in the room :P

0celo7

You don't know?

Slereah

Damn Windows always changing the interface

yuggib

Josip Broz Tito

Josip Broz Tito (Cyrillic: Јосип Броз Тито, pronounced [jǒsip brôːz tîto]; born Josip Broz; 7 May 1892 – 4 May 1980) was a Yugoslav revolutionary and statesman, serving in various roles from 1943 until his death in 1980. During World War II he was the leader of the Partisans, often regarded as the most effective resistance movement in occupied Europe. While his presidency has been criticized as authoritarian, Tito was "seen by most as a benevolent dictator" due to his economic and diplomatic policies. He was a popular public figure both in Yugoslavia and abroad. Viewed as a unifying symbol, his...

Slereah

15:12

GeForce GTX 980

yuggib

yugoslavia died with him...

ACuriousMind

@Slereah Press Windows+R, type dxdiag, fastest way to get specs.

0celo7

@ACuriousMind What is the condition for two lines of slopes $m$ and $m'$ to be orthogonal

@Slereah Jesus Christ

You have the second most powerful GPU on the market and you don't even know what it is?

Slereah

I got the new PC for my birthday B)

0celo7

Custom or ordered?

Slereah

15:14

Assembled by a shop

Also this is 2015

Danu

What are the one-point compactifications of $\mathbb{R}^*$ and $\mathbb{C}^*$?

Bastian Treichler

@ACuriousMind you get a notification if you're not in the chat. so we just spammed all 63k users :)

Slereah

I haven't cared about the specs of my PC in a while

Reals is the circle, complex is the Riemann sphere

In the 90's I cared a lot about the specs because that was the difference between a game running and not running

ACuriousMind

@Danu You really need to stop using your personal commands in MathJax :P

Slereah

But nowadays?

ACuriousMind

15:15

Also, what's the asterisk?

0celo7

@ACuriousMind pls

Slereah

Who cares

Sean

@ACuriousMind Don't make me do the right hand rule! haha

0celo7

What is $S^1\times S^2$

Slereah

a sphere running around in a circle

Weee

ACuriousMind

15:16

@0celo7 What? I'm not telling you something you can easily check yourself, you need to develop some confidence.

Slereah

Console people like to mock PC people for crashes and all

0celo7

@ACuriousMind I know it's $m=-1/\tilde m$

Danu

@ACuriousMind I think someone should change those commands to conform with my standards. Also, the star denotes the multiplicative group.

0celo7

So I really have no clue what I did wrong

Slereah

But really it's been like ten years since I've had any regular game crashes

yuggib

15:17

@Danu Are them not the projective spaces?

0celo7

@Slereah You played Skyrim, right?

Danu

@yuggib Oh... Could be

ACuriousMind

@Danu Ew, that should be $\mathbb{C}^\times$, not a star. :P

Slereah

Yes.

Danu

@ACuriousMind WRONG :P

Slereah

15:17

I've had bugs in Skyrim

But crashes?

0celo7

Skyrim didn't crash for you

Danu

I know you use that notation

but it 's wrong.

Slereah

Pretty rare

0celo7

Bull

Did you have mods?

Danu

Note the typesetting inconvenience :P

Slereah

15:18

Plenty

0celo7

How is that even possible

Slereah

It's all in the reflexes

0celo7

Skyrim + mods = one giant crash

@Slereah wtf

Slereah

(Also saying that movie reference to a 17 years old feels like a waste)

yuggib

@Danu That's almost the message of God to his creation

Danu

15:18

Who denotes multiplication by a $\times$ anyways, except elementary school teachers?

0celo7

@ACuriousMind do you think Skyrim crashes a lot?

@Slereah no 17 year olds in here

@Danu ...

I do

you're telling me you do $6.67\cdot10^{-11}$?

Danu

@0celo7 Lol

yuggib

@0celo7 It has been a while I had not to write a number

Danu

I should've added elementary school kids to that group.

ACuriousMind

@yuggib No, that would just be the one-point compactification of $\mathbb{R}$

Danu

15:21

@0celo7 Yes

@ACuriousMind ? No, that'd be $S^1$

0celo7

@Danu That's wrong.

Danu

@0celo7 Sure :)

0celo7

@Danu It is.

Don't worry.

ACuriousMind

@Danu Oh, yuggib meant that $\mathbb{R}^*$ is a projective space, not the one-point compactification of it?

Danu

Cool story :)

0celo7

15:22

You can fix your notation now, I won't be a dick about it.

Unlike some people and quadratic equations.

ACuriousMind

Anyway, that's wrong too because the real projective line is just $S^1$.

Danu

@ACuriousMind I think he meant the 1pc of it is, right @yuggib?

@ACuriousMind No antipodal involution?

yuggib

@Danu Yes, but maybe I misunderstood what $\mathbb{R}^*$ is...

ACuriousMind

@Danu No.

That happens for the real projective plane, but not for the line

I.e. $\mathbb{R}P^2$ is such a weird quotient of $S^2$, but $\mathbb{R}P^1$ is just $S^1$.

Danu

@ACuriousMind Sorry, what does your terminology mean exactly? AFAIK the real projective line is the space of all 1-dim subspaces of $\mathbb{R}^2$

...which strongly suggests antipodal involution

0celo7

15:24

you have to eliminate k

ACuriousMind

@Danu Yes,

0celo7

somehow

ACuriousMind

Ah

Danu

...right.

ACuriousMind

Well, there is antipodal involution, but $S^1/\mathbb{Z}_2 = S^1$ :D

Danu

15:25

@ACuriousMind Okay, fine :D

My homie here suggests it's the figure-eight.

the 1pc of $\mathbb{R}^*$, that is

0celo7

lol

Danu

damnit tex

ACuriousMind

Just draw it carefully, you see that $S^1/\mathbb{Z}_2$ is really just a "smaller" circle.

Danu

$\newcommand{\Reals}{\mathbb{R}} \newcommand{\Complexes}{\mathbb{C}}$

woop

0celo7

@ACuriousMind Is that @ me? I took it back because I got it.

Danu

15:26

$\Reals$

perfect

0celo7

@Danu you broke the TeX

Danu

@ACuriousMind Yeah, sure

@0celo7 I ain't breaking nuttin'

Slereah

Oh buns

ACuriousMind

@Danu He's correct.

Slereah

I saved just before a grenade strike that kills me

0celo7

15:27

Actually, you just broke the chat.

Danu

and for $\Complexes$, the complex figure eight????

What does that even mean :P

@0celo7 Working fine over here :)

0celo7

Figure eight in complex numbers

Maybe you should go back to the basics

Quadratic equations, etc.

Danu

^ no u

ACuriousMind

The topology of $\mathbb{R}^*$ is $\mathbb{R}\cup\mathbb{R}$, and compactifying each part gives an $S^1$. Since we only add one point, we must add the same both times, this gives $S^1 \vee S^1$ for $\vee$ the wedge sum.

0celo7

Dumb question, $\mathbb{R}\cup\mathbb{R}=\mathbb{R}^2$?

Danu

15:29

No

Try proving it ;D

0celo7

wtf is the smash product

@Danu trying to figure out caclulus

Danu

^already did

0celo7

I really am terrible at math

ACuriousMind

Dammit, I meant the wedge sum, not the smash product

0celo7

what is the wedge sum

Danu

15:30

Mah homie says it's the connected sum

ACuriousMind

@Danu Connected sum means cutting little disks out and gluing, the connected sum of $S^1$ and $S^1$ is $S^1$ again.

BTW, $S^1\vee S^1$ is the figure-eight.

Danu

@ACuriousMind Coolbeans

The wedge sum really seems a bit overkill, terminology-wise, or is it very important in a certain setting?

0celo7

No one is going to answer my questions

ACuriousMind

Now, the one-point compactification of $\mathbb{C}^*$ is a bit weirder - it's the sphere where the north and southpole have been identified

0celo7

I'm like a fly on the wall, buzzing

Danu

15:32

@ACuriousMind That makes a lot of sense

It looks like a torus

A "degenerate torus"

Sean

@ACuriousMind so I just used a strong electromagnet and it definitely looks like my test bar magnet is mislabled. so why would that be?

ACuriousMind

@Danu It's quite important in the setting of homotopy groups to formally define a bouquet of circles.

Danu

@ACuriousMind Lel

Slereah

But $R \cup R$ is just $R$ :O

ACuriousMind

@Slereah I meant a disjoint union

Danu

15:34

@Slereah take the square cup

Slereah

Well WRITE IT THEN

Danu

or the one with a dot

ACuriousMind

@Danu Yep

0celo7

What is a disjoint union, anyway

Danu

@0celo7 One that doesn't intersect

0celo7

15:35

I know that

Danu

...

0celo7

So what is the disjoint union of two Rs

ACuriousMind

Nothing, it's just their disjoint union. It doesn't have a special name

0celo7

What would a picture of it be

ACuriousMind

Two lines.

(not touching)

0celo7

15:37

Dude that's totally R^2

ACuriousMind

Uh, no.

yuggib

still can't understand that weird notation $\mathbb{R}^*$...

Slereah

Disjoint union is fancy talk for "Ordered pair"

0celo7

Exactly!

Ordered pair of two reals is R2

yuggib

@Slereah not exactly

at least, it depends what do you mean by ordered pair

ACuriousMind

15:38

@yuggib It comes from ring theory, for a ring $R$, $R^*$ is the set of all invertible elements. For fields like $\mathbb{R},\mathbb{C}$, it's just fancy-speak for "without zero".

Danu

@0celo7 lol

Slereah

We should have some SE entry for the definition of "fancy talk"

Danu

@yuggib $\Reals$ with a multiplication $*$, i.e. the multiplicative part of $\Reals$

Makes sense, no?

yuggib

@Danu no it doesn't

ACuriousMind

@Slereah I am more used to the idea that the Cartesian product of two sets is the set of ordered pairs, which is larger than the disjoint union.

0celo7

15:39

So I guess it's funny that I don't get this and not worth explaining

Danu

@yuggib citation needed

@ACuriousMind This

yuggib

$\mathbb{R}$ already has a multiplication...if you want it to be invertible for any element is all another thing

so it does not make so much sense for me

Danu

@yuggib group

ACuriousMind

@0celo7 I don't see what there is to explain - two lines are not a plane.

Danu

I forgot that word earlier

So yeah

I guess it's not super

"water tight"

interpretation-wise

0celo7

15:41

@ACuriousMind Of course it is. Points in R2 can be defined by projections onto two lines.

Danu

@0celo7 So?

ACuriousMind

@0celo7 That is not at all the same.

Danu

That's like saying that the two axes of $\Reals^2$ are $\Reals^2$

0celo7

I think they are.

Danu

Then try to construct a homeomorphism.

0celo7

15:42

That's what Cartesian coordinates are

ACuriousMind

"Two lines" means the space is really just two lines, i.e. just the points on the two lines. This is very different from saying a point in the plane is labeled by one point on each axis.

yuggib

nevertheless, I am happy that you and @ACuriousMind understood each other, still the question you asked seems a little bit ambiguous without stating before which topology to take for $\mathbb{R}^*$ in the beginning

Danu

@yuggib Always the subspace topology

ACuriousMind

@yuggib It's implicit to take the subspace topology

Danu

(the obvious one)

when not stated explicitly

I don't believe that anyone explicitly writes out when they're taking the subspace topology in any book.

0celo7

15:43

@ACuriousMind Sure, but there's a nice map into the plane given by Cartesian coordinates.

I'm not trolling, I really don't see what you're trying to tell me here.

yuggib

@Danu The subspace topology of what? of $\mathbb{R}^*$ as a subset of $\mathbb{R}$?

ACuriousMind

@0celo7 There is no homeomorphism between $\mathbb{R}\cup \mathbb{R}$ and $\mathbb{R}\times \mathbb{R}$. See, an element of $\mathbb{R}\cup\mathbb{R}$ is just a real number $x$ together with a label $i\in\{1,2\}$ stating to which of the lines it belongs. An element in $\mathbb{R}^2$ is a pair of real numbers $(x,y)$.

yuggib

why not as a subspace of $\aleph_{2^{\aleph_0}}$...

ACuriousMind

(One can construct a crazy bijection, but that's because $\mathbb{R}$ and $\mathbb{R}^n$ have the same cardinality and that should not concern us here.)

Danu

@yuggib Because we're not doing set theory

@ACuriousMind The bijection is not that crazy. It can even be continuous.

@yuggib I honestly think there's at least a 95% chance of correct interpretation (after explaining what the $*$ means) if you ask mathematicians

ACuriousMind

15:49

@Danu From $\mathbb{R}$ to $\mathbb{R}^n$ I'll grant that, but not from $\mathbb{R}\cup\mathbb{R}$.

Danu

@ACuriousMind Hmkay :)

yuggib

@Danu Well...I would not be so sure about that

0celo7

@ACuriousMind Wait, why does it have to be the same number on both lines

Or is $x_1\ne x_2$

Danu

@0celo7 There are no two points for $\Reals\cup \Reals$

0celo7

TeX

Danu

15:51

(still working for me)

You first pick one of the two copies, then specify where on it

this is a prescription to specify any element

i.e. $(1, x\in \Reals)$

or $(2,y\in\Reals)$

0celo7

@yuggib the heck is that

::looks up definition of union::

Hmm

Why didn't you just tell me that.

Moving on.

yuggib

@0celo7 An aleph, i.e. a cardinal number

0celo7

@yuggib What's that?

yuggib

@0celo7 a number that counts the cardinality of a set

0celo7

@yuggib What's that?

yuggib

15:53

i.e. the number of elements in it

0celo7

What's the cardinality of the reals?

yuggib

@0celo7 $2^{\aleph_0}$, where $\aleph_0$ is the cardinality of $\mathbb{N}$

i.e. the first infinite cardinal

0celo7

Brain melted.

Danu

Continuum hypothesis, true or false? Votes ready? GO!

True

(no nitpicking and subtleties allowed)

0celo7

tralse

yuggib

15:56

there are models where it is true, and models where it is false

within ZFC

so it is not so important

Danu

40 secs ago, by Danu

(no nitpicking and subtleties allowed)

yuggib

however I would vote for false

0celo7

this chat is a set theory chat whenever @yuggib comes in

ACuriousMind

Sep 8 at 22:06, by yuggib

A completely random and unrelated question (since the majority of the live ones are mathematicians): do you think that the continuum hypothesis is true?

Sep 8 at 22:08, by Danu

It doesn't really matter... As far as I can tell

yuggib

:-D

ACuriousMind

15:57

We've had this discussion before

Danu

@ACuriousMind :D

that's great

yuggib

the inversion of roles

Danu

10/10

0celo7

I'm not even going to bother asking what the continuum hypothesis is.

But I'm sure German middle schoolers know

yuggib

$\aleph_1=2^{\aleph_0}$

Danu

15:58

ACM's chat memory is the best thing in this entire chat.

0celo7

So I'm just going to leave

Danu

@0celo7 The ol' "screw you guys, I'm going home"?

ACuriousMind

@Danu @ChrisWhite's memory rivals mine

Danu

@ACuriousMind Well he ain't here so often.

yuggib

the generalized continuum hypothesis is that for any ordinal $\alpha$, $\aleph_{\alpha+1}=2^{\aleph_\alpha}$

0celo7

15:59

@Danu The ol' I can't figure out this test problem no one wants to help me and people laugh at me and are too smart and I'm going to class

close, but no cigar

Danu

@0celo7 Don't take everything so harshly

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